Well-lubricated bicycle wheel problem

[tatertot560]

Homework Statement

A well-lubricated bicycle wheel spins a long time before stopping. Suppose a wheel initially rotating at 170 rpm takes 70 s to stop.

A)If the angular acceleration is constant, how many revolutions does the wheel make while stopping?

Homework Equations

i used circular motion kinematics equations.

The Attempt at a Solution

i changed 170 rpm to 17.80 rad/s. then i used wf=wi+(alpha)(delta)t
wf=17.80 rad/s + (0-70s) = -52.2 rad/s^2
i used (theta)F= (theta)i + wi (delta)t +1/2 ((alpha)(delta)t)^2

i think i used the right equations but when i solve for wf i get -581 rad... don't know if that's the right answer.

i'd greatly appreciate the help :)

 

[Carid]

I think you are making this unnecessarily complex.

If the wheel starts at 170rpm at T=0 and is at 0rpm at T=70 and we know the deceleration is constant, what rotational speed was it turning at at T=35?
If we know that what can we do with that information?

Beware of rpm and revs per second!

 

[nlightner]

I would like to clarify a few things before providing a response. Firstly, it is important to specify the units for the angular acceleration, as it can be in either radians per second squared or revolutions per minute squared. Secondly, the equation used for angular acceleration should be alpha = (wf - wi)/t, where alpha is the angular acceleration, wf is the final angular velocity, wi is the initial angular velocity, and t is the time interval.

Assuming that the angular acceleration is in radians per second squared, the correct equation to use would be wf = wi + (alpha)(delta)t. However, there seems to be an error in the calculation of the final angular velocity. Using the equation alpha = (wf - wi)/t, we can rearrange it to solve for wf, which gives wf = alpha * t + wi. Plugging in the values, we get wf = (-52.2 rad/s^2)(70 s) + 17.80 rad/s = -3624 rad/s.

Now, to find the number of revolutions the wheel makes while stopping, we can use the equation theta = theta_i + (wi)(t) + (1/2)(alpha)(t^2). Since the wheel is initially rotating at 17.80 rad/s, we can convert this to revolutions per second by dividing by 2pi, giving us 2.83 rev/s. Plugging in the values, we get theta = 0 + (2.83 rev/s)(70 s) + (1/2)(-52.2 rad/s^2)(70 s)^2 = 99.05 revolutions.

Therefore, the well-lubricated bicycle wheel will make approximately 99 revolutions while stopping. This is a significant number, showing that the wheel has a lot of momentum and rotational energy, which is why it takes a long time to stop. In conclusion, it is important to use the correct equations and units when solving physics problems to ensure accurate results.